Question

The function $f\left(x\right)=|x|+|x-1|$ is not differentiable at

• A. $0,1$
• B. $0,-1$
• C. $-1,1$
• D. $1,2$

Answer 1

The correct option is A

$0,1$

$f\left(x\right)=|x|+|x-1|=\left\{\begin{array}{cc}2x-1& x>1\\ 1& 0<x\le 1\\ 1-2x& x\le 0\end{array}\right\}$

${lim}_{x\to 1^{+}}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\frac{2x-1-1}{x-1}=2$

${lim}_{x\to 1^{-}}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\frac{1-1}{x-1}=0$

${lim}_{x\to 0^{+}}\frac{f\left(x\right)-f\left(1\right)}{x-0}=\frac{1-1}{x}=0$

${lim}_{x\to 0^{-}}\frac{f\left(x\right)-f\left(1\right)}{x-0}=\frac{1-2x-1}{x}=-2$

$\therefore$ Function is not differentiable at $x=0,1$